XQUERY return 结果格式 <!ELEMENT dynosaur ($dynosaurName) +>

Question

我在 .xml 文件中使用 XQUERY FLWOR 表达式：

for $dynosaur in doc("document.xml")//species
 let $dynosaurName := $dynosaur/text()        // keeping the dynosaurName as a variable
return $dynosaur

以上 return 结果如下：

<species age="84">Velociraptor</species>

我需要格式化结果如下：

<!ELEMENT dynosaur (Velociraptor) +>

所以我正在尝试使用以下但不起作用...

return <!ELEMENT dynosaur ({$dynosaurName}) +> //here i want that format but it return error

和xml文件：

<?xml version="1.0"?>
<dinosauria>
    <group>
        <name>saurishia</name>
        <subgroups>
            <group>
                <name>theropoda</name>
                <subgroups>
                    <group>
                        <name>carnosaurs</name>
                        <speciesList>
                            <species age="74">Dyptosaurus</species>
                            <species age="170">Megalosaurus</species>
                            <species age="67">Tyrannosaurus</species>
                        </speciesList>
                    </group>
                    <group>
                        <name>coelurosauria</name>
                        <speciesList>
                            <species age="84">Velociraptor</species>
                            <species age="110">Deinonychus</species>
                            <species age="228">Eoraptor</species>
                        </speciesList>
                    </group>
                </subgroups>
            </group>
            <group>
                <name>sauropodomorpha</name>
                <subgroups>
                    <group>
                        <name>sauropods</name>
                        <speciesList>
                            <species age="155">Brachiosaurus</species>
                            <species age="155">Camarasaurus</species>
                        </speciesList>
                    </group>
                </subgroups>
            </group>
        </subgroups>
    </group>
    <group>
        <name>ornithishia</name>
        <subgroups></subgroups>
    </group>
</dinosauria>

最后：

我找不到任何方法来return那种类型的结果。考虑到这本书，我检查了很多链接：http://www.datypic.com/books/xquery/chapter09.html

Answer 1

你可以尝试输出文本吗？

return concat("<!ELEMENT dynosaur (",$dynosaurName, ") +>")

XQUERY return 结果格式 <!ELEMENT dynosaur ($dynosaurName) +>

XQUERY return result in format <!ELEMENT dynosaur ($dynosaurName) +>

xpath

xquery

flwor